Given below are two statements :
Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.
Statement II : The time period of revolution of the satellite is T=2π√(Rₑ/g) (for satellite very close to the earth surface), where Rₑ radius of earth and g acceleration due to gravity. In the light of the above statements, choose the correct answer from the options given below :

Solution

Physics sir · 4 steps

Write g in terms of density and the radius cancels, leaving T depending on density alone.

1. Period near surface. For an orbit just above the surface, gravity supplies the centripetal force with r ≈ Rₑ, so mg = mω² Rₑ. With ω = 2π/T this gives T = 2π√(Rₑ/g), so Statement II is true.
2. Bring in density. Write g = GM/Rₑ² with M = 4/3π Rₑ³ ρ. Then g = 4/3π G ρ Rₑ.

   T = 2π√(Rₑ/(4/3π Gρ Rₑ)) = √(3π/Gρ)

3. Density decides. Rₑ cancels, and T depends only on the density ρ of the earth. So Statement I is also true.
4. So. Both statements are true, which is option (c).

The idea to keep. For a satellite skimming any planet, T = √(3π/Gρ). Size drops out, so only the planet's density sets the period.

Answer from the NTA final answer key. Solution by PrepWiser.